Solve the system of equations by substitution, if a solution exists.
-$$\left\{ \begin{array} { l }
x + 2 y = 16 \
2 x + 3 y = 24
\end{array} \right.$$
A) $x = - 8 , y = 0$
B) No solution
C) $x = 0 , y = 8$
D) $x = 1 , y = 7$

Question

Solve the system of equations by substitution, if a solution exists.
-$$\left\{ \begin{array} { l } 
x + 2 y = 16 \
2 x + 3 y = 24
\end{array} \right.$$
A) $x = - 8 , y = 0$
B) No solution
C) $x = 0 , y = 8$
D) $x = 1 , y = 7$

Quizplus · Accepted Answer

Solving the first equation for $x$, we get $x = 16 - 2y$. Substituting this into the second equation gives $2(16 - 2y) + 3y = 24$, which simplifies to $32 - 4y + 3y = 24$, and then to $32 - y = 24$, leading to $y = 8$. Substituting $y = 8$ back into $x = 16 - 2y$ gives $x = 16 - 2(8) = 0$.

Solve the System of Equations by Substitution, If a Solution {x+2y=162x+3y=24\left\{ \begin{array} { l } x + 2 y = 16 \\2 x + 3 y = 24\end{array} \right.{x+2y=162x+3y=24​

Solve the System of Equations by Substitution, If a Solution $\left\{ \begin{array} { l } x + 2 y = 16 \\2 x + 3 y = 24\end{array} \right.$